Sec 3 A-Math Polynomials, cubic equations and partial fractions worksheet with answers
A free Sec 3 A-Math worksheet on polynomials, cubic equations and partial fractions: 7 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.
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Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.
Section A
1
(a)Find the quotient when x3+3x2−5x−15 is divided by x+3.
Answer (a)[1]
(b)Find the quotient when x3+x2−5x−2 is divided by x−2.
Answer (b)[1]
2
Solve x3−6x2+5x+12=0.
Answer[3]
3
(a)Find the remainder when 2x3+x2−4x+7 is divided by x−1.
Answer (a)[2]
(b)Given that x+1 is a factor of x3+3x2−2x+k, find the value of k.
Answer (b)[2]
4
(a)Find the remainder when x3−2x2+5x−3 is divided by 3x−2.
Answer (a)[2]
(b)Find the remainder when x3−2x2+3x is divided by x2−4.
Answer (b)[3]
5
(a)Express (x−3)(x+1)2x2+3x−11 in partial fractions.
Answer (a)[3]
(b)Express (x+2)(x−1)2x2−x−13 in partial fractions.
Answer (b)[3]
Section B
6
It is given that f(x)=x3+ax2−10x−24, where a is a constant, and that x−3 is a factor of f(x).
(a)Find the value of a.
Answer (a)[2]
(b)Factorise f(x) completely.
Answer (b)[3]
(c)Solve the equation f(x)=0.
Answer (c)[1]
(d)Find the remainder when f(x) is divided by x+1.
Answer (d)[2]
7
(a)Find the remainder when x3+3x2−2x+1 is divided by x+1.
Answer (a)[3]
(b)Solve x3−9x2+22x−12=0, expressing non-integer roots in surd form.
Answer (b)[5]
(c)Express (x−1)(x−3)−2x2+10x−14 in partial fractions.
A root: x=−1f(−1)=(−1)3−6(−1)2+5(−1)+12=−1−6−5+12=0So (x+1) is a factor.x3−6x2+5x+12=(x+1)(x2+px+q)(x+1)(x2+px+q)=x3+px2+qx+x2+px+q=x3+(p+1)x2+(q+p)x+qp+1=−6p=−7q=12Check the x term: q+p=12+(−7)q+p=5 ✓x3−6x2+5x+12=(x+1)(x−3)(x−4)x=−1x=3x=4
Question 3
(a)Answer: The remainder is 6.[2]
x−1=0 when x=1f(1)=2(1)3+(1)2−4(1)+7=2+1−4+7=6The remainder is 6.
(b)Answer: k=−4[2]
f(−1)=0(−1)3+3(−1)2−2(−1)+k=0k−1+3+2=0k+4=0k=−4
Question 4
(a)Answer: The remainder is −277.[2]
3x−2=0 when x=32f(32)=(32)3−2(32)2+5(32)−3=278+9−8+310−3=−277The remainder is −277.
x+1=0 when x=−1f(−1)=(−1)3+3(−1)2−2(−1)+1=−1+3+2+1=5The remainder is 5.
(b)Answer: x=3−5 or x=3+5[5]
A root: x=3f(3)=(3)3−9(3)2+22(3)−12=27−81+66−12=0So (x−3) is a factor.x3−9x2+22x−12=(x−3)(x2+px+q)(x−3)(x2+px+q)=x3+px2+qx−3x2−3px−3q=x3+(p−3)x2+(q−3p)x−3qp−3=−9p=−6−3q=−12q=4Check the x term: q−3p=4−3×(−6)q−3p=22 ✓x=2(1)−(−6)±(−6)2−4(1)(4)=26±20x=3x=3−5 or x=3+5